¡¾ÌâÄ¿¡¿ÒÀ¾ÝͼÖеªÔªËؼ°Æ仯ºÏÎïµÄת»¯¹Øϵ£¬»Ø´ðÎÊÌ⣺

£¨1£©ÊµÑéÊÒ³£ÓÃNH4ClÓëCa(OH)2ÖÆÈ¡°±Æø£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________¡£

£¨2£©ÈôÒªÊÕ¼¯Ò»Æ¿°±Æø£¬ÇëÔÚͼÐé¿òÄÚ»­³öÁ¬½Óͼ_____¡£

£¨3£©¹¤ÒµÉÏÒÔNH3¡¢¿ÕÆø¡¢Ë®ÎªÔ­ÁÏÉú²úÏõËáµÄ¹¤ÒÕÁ÷³Ì¼òͼÈçÏÂËùʾ£º

д³öNH3¡úNOµÄ»¯Ñ§·½³Ìʽ__________________________________¡£

£¨4£©Í¼ÖУ¬XµÄ»¯Ñ§Ê½Îª_______£¬½«Xת»¯ÎªHNO3ÊôÓÚ______________·´Ó¦£¨Ìî¡°Ñõ»¯»¹Ô­¡±»ò¡°·ÇÑõ»¯»¹Ô­¡±£©·´Ó¦¡£

£¨5£©ÈôÒª½«NH3¡úN2£¬´ÓÔ­ÀíÉÏ¿´£¬ÏÂÁÐÊÔ¼Á¿ÉÐеÄÊÇ________£¨ÌîÐòºÅ£©¡£

A£®O2 B£®Na C£®NH4Cl D£®NO2

¡¾´ð°¸¡¿2NH4Cl£«Ca(OH)22NH3¡ü£«CaCl2£«2H2O 4NH3£«5O24NO£«6H2O N2O5 ·ÇÑõ»¯»¹Ô­ AD

¡¾½âÎö¡¿

£¨1£©ÊµÑéÊÒ³£ÓÃNH4ClÓëCa(OH)2ÖÆÈ¡°±Æø£¬Éú³ÉÂÈ»¯¸Æ¡¢°±ÆøºÍË®£»

£¨2£©°±Æø¼«Ò×ÈÜÓÚË®£¬±È¿ÕÆøÇᣬÊÕ¼¯·½·¨Ö»ÄÜÓÃÏòÏÂÅÅ¿ÕÆø·¨ÊÕ¼¯£¬µ¼Æø¹ÜλÖö̽ø³¤³ö£»

£¨3£©°±ÆøµÄ´ß»¯Ñõ»¯£º4NH3+5O24NO+6H2O£»¢ÚNO¡úNO2ʵÑéÏÖÏóÊÇÎÞÉ«ÆøÌå±ä»¯Îªºì×ØÉ«ÆøÌ壻¢Û¶þÑõ»¯µªºÍË®·´Ó¦Éú³ÉÒ»Ñõ»¯µªºÍÏõË᣻

£¨4£©Í¼1·ÖÎö¿ÉÖªX»¯ºÏ¼ÛÓëÏõËáÏàͬΪ+5¼Û£¬ÎïÖÊÀàÐÍΪÑõ»¯ÎX»¯Ñ§Ê½Îª£ºN2O5£¬´ÓÎïÖÊÐÔÖÊÉÏ¿´£¬XÊôÓÚËáÐÔÑõ»¯ÎïN2O5+2H2O=2HNO3£¬×ª»¯ÎªHNO3ÊôÓÚ·ÇÑõ»¯»¹Ô­·´Ó¦£»

£¨7£©ÈôÒª½«NH3¡úN2,´ÓÔ­ÀíÉÏ¿´£¬¾ßÓÐÑõ»¯ÐÔµÄÊÔ¼Á¿ÉÐУ»

£¨1£©ÊµÑéÊÒ³£ÓÃNH4ClÓëCa(OH)2ÖÆÈ¡°±Æø£¬Éú³ÉÂÈ»¯¸Æ¡¢°±ÆøºÍË®£¬»¯Ñ§·½³ÌʽΪ2NH4Cl+Ca(OH)2CaCl2+2NH3+2H2O£»

ÕýÈ·´ð°¸£º2NH4Cl£«Ca(OH)22NH3¡ü£«CaCl2£«2H2O

£¨2£©°±Æø¼«Ò×ÈÜÓÚË®£¬±È¿ÕÆøÇᣬÊÕ¼¯·½·¨Ö»ÄÜÓÃÏòÏÂÅÅ¿ÕÆø·¨ÊÕ¼¯£¬µ¼Æø¹ÜλÖö̽ø³¤³ö£»£»

ÕýÈ·´ð°¸£º

£¨3£©°±ÆøµÄ´ß»¯Ñõ»¯£º4NH3+5O24NO+6H2O£»¢ÚNO¡úNO2ʵÑéÏÖÏóÊÇÎÞÉ«ÆøÌå±ä»¯Îªºì×ØÉ«ÆøÌ壻¢Û¶þÑõ»¯µªºÍË®·´Ó¦Éú³ÉÒ»Ñõ»¯µªºÍÏõËᣬ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º3NO2+H2O=2HNO3+NO£»

ÕýÈ·´ð°¸£º4NH3+5O24NO+6H2O¡£

£¨4£©Í¼1·ÖÎö¿ÉÖªX»¯ºÏ¼ÛÓëÏõËáÏàͬΪ+5¼Û£¬ÎïÖÊÀàÐÍΪÑõ»¯ÎX»¯Ñ§Ê½Îª£ºN2O5£¬´ÓÎïÖÊÐÔÖÊÉÏ¿´£¬XÊôÓÚËáÐÔÑõ»¯ÎïN2O5+2H2O=2HNO3£¬×ª»¯ÎªHNO3ÊôÓÚ·ÇÑõ»¯»¹Ô­·´Ó¦£»

ÕýÈ·´ð°¸£ºN2O5 ·ÇÑõ»¯»¹Ô­¡£

£¨7£©ÈôÒª½«NH3¡úN2,´ÓÔ­ÀíÉÏ¿´£¬¾ßÓÐÑõ»¯ÐÔµÄÊÔ¼Á¿ÉÐУ»

A.O2¾ßÓÐÑõ»¯ÐÔ£¬¹ÊAÕýÈ·£»

B.NaÖ»¾ßÓл¹Ô­ÐÔ£¬¹ÊB´íÎó£»

C.NH4ClÓë°±Æø²»·´Ó¦£¬¹ÊC´íÎó£»

D.NO2¾ßÓÐÑõ»¯ÐÔ£¬¹ÊDÕýÈ·£»

ÕýÈ·´ð°¸£ºAD

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©K2Cr2O7 + 14HCl= 2KCl + 2CrCl3 + 3Cl2¡ü+ 7H2O £¨Óá°µ¥ÏßÇÅ¡±±íʾµç×ÓתÒƵķ½ÏòºÍÊýÄ¿£©___,Ñõ»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ_____________¡£

£¨2£©______mol H2OÖй²º¬ÓÐ9.03¡Á1022¸öÔ­×Ó£¬ÆäÖÊÁ¿Îª_______¡£

£¨3£©ÅäƽÏÂÁÐÑõ»¯»¹Ô­·´Ó¦·½³Ìʽ£º___KMnO4+___H2S+__H2SO4(Ï¡) ¡ª__MnSO4+__S¡ý+__K2SO4+__H2O

£¨4£©Cl2ÊÇÒ»ÖÖÓж¾ÆøÌ壬Èç¹ûй©»áÔì³ÉÑÏÖصĻ·¾³ÎÛȾ¡£»¯¹¤³§¿ÉÓÃŨ°±Ë®À´¼ìÑéCl2ÊÇ·ñй©£¬Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º3Cl2£¨Æø£©£«8NH3£¨Æø£©£½6NH4Cl£¨¹Ì£©£«N2£¨Æø£©£¬Èô·´Ó¦ÖÐÏûºÄCl2 1.5 mol, Ôò±»Ñõ»¯µÄNH3ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ______ L¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁгýÔÓËùÑ¡ÓõÄÊÔ¼Á¼°²Ù×÷·½·¨¾ùÕýÈ·µÄÒ»×éÊÇ(À¨ºÅÄÚΪÔÓÖÊ)

Ñ¡Ïî

´ýÌá´¿µÄÎïÖÊ

Ñ¡ÓõÄÊÔ¼Á

²Ù×÷·½·¨

A

NaHCO3(Na2CO3)

ÊÊÁ¿ÑÎËá

Õô·¢½á¾§

B

CO2(CO)

O2

µãȼ

C

Mg(Al)

ÇâÑõ»¯ÄÆÈÜÒº

¹ýÂË

D

CO2(HCl)

ÇâÑõ»¯ÄÆÈÜÒº

Ï´Æø

A. A B. B C. C D. D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³ÈÜÒºÓÉNa+¡¢Cu2+¡¢Ba2+¡¢Fe3+¡¢AlO¡¢CO¡¢SO¡¢ClÖеÄÈô¸ÉÖÖÀë×Ó×é³É£¬È¡ÊÊÁ¿¸ÃÈÜÒº½øÐÐÈçÏÂʵÑ飺ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. Ô­ÈÜÒºÖÐÒ»¶¨Ö»´æÔÚ¡¢¡¢¡¢Cl-ËÄÖÖÀë×Ó

B. ÆøÌåAµÄ»¯Ñ§Ê½ÊÇCO2£¬Æäµç×ÓʽΪO::C::O

C. Ô­ÈÜÒºÖÐÒ»¶¨²»´æÔÚµÄÀë×ÓÊÇCu2+¡¢Ba2+¡¢Fe3+

D. Éú³É³ÁµíBµÄÀë×Ó·½³ÌʽΪ£ºAl3++3OH===Al(OH)3¡ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿NA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÊýÖµ£¬ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ

A. ³£Î³£Ñ¹Ï£¬1.7 g°±ÆøÖк¬ÓеÄÔ­×ÓÊýĿΪ0.4NA

B. 50 mL 1 mol¡¤L1 K2SO4ÈÜÒºÖк¬ÓеÄK+ÊýĿΪ0.1NA

C. 5.6 gÌúÓë×ãÁ¿Ï¡ÁòËᷴӦתÒƵĵç×ÓÊýΪ0.3NA

D. ±ê×¼×´¿öÏ£¬4.48 LµÄÑõÆøºÍµªÆøµÄ»ìºÏÎﺬÓеķÖ×ÓÊýĿΪ0.2NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»¯Ñ§¼ÆÁ¿ÔÚ»¯Ñ§ÖÐÕ¼ÓÐÖØÒªµØλ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©0.3 mol NH3·Ö×ÓÖÐËùº¬Ô­×ÓÊýÓëÔ¼__________¸öH2O·Ö×ÓÖÐËùº¬Ô­×ÓÊýÏàµÈ¡£

£¨2£©V mLº¬a g Al3+µÄAl2(SO4)3ÈÜÒºÖÐËùº¬SO42µÄÎïÖʵÄÁ¿Å¨¶ÈΪ_____ mol¡¤L1¡£

£¨3£©ÔÚÒ»¶¨Î¶ȺÍѹǿÏ£¬1Ìå»ýÆøÌåX2Óë3Ìå»ýY2»¯ºÏÉú³É2Ìå»ýÆø̬»¯ºÏÎ¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½Îª_________¡£

£¨4£©½«¸÷0.3 molµÄÄÆ¡¢Ã¾¡¢ÂÁ·Ö±ð·ÅÈë100 mL 1 mol¡¤L1µÄÑÎËáÖУ¬Í¬ÎÂͬѹϲúÉúÆøÌåµÄÌå»ý±ÈΪ_____________¡£

£¨5£©Ð¿ÓëºÜÏ¡µÄÏõËá·´Ó¦Éú³ÉÏõËáп¡¢ÏõËá狀ÍË®£¬µ±Éú³É1 molÏõËáпʱ£¬²Î¼Ó·´Ó¦µÄÏõËáµÄÎïÖʵÄÁ¿Îª_________ mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Î¢ÉúÎïµç³ØÊÇÖ¸ÔÚ΢ÉúÎïµÄ×÷ÓÃϽ«»¯Ñ§ÄÜת»¯ÎªµçÄܵÄ×°Öã¬Æ乤×÷Ô­ÀíÈçͼËùʾ¡£ÏÂÁÐÓйØ΢ÉúÎïµç³ØµÄ˵·¨´íÎóµÄÊÇ ( )

A. Õý¼«·´Ó¦ÖÐÓÐCO2Éú³É

B. ΢ÉúÎï´Ù½øÁË·´Ó¦Öеç×ÓµÄתÒÆ

C. ÖÊ×Óͨ¹ý½»»»Ä¤´Ó¸º¼«ÇøÒÆÏòÕý¼«Çø

D. µç³Ø×Ü·´Ó¦ÎªC6H12O6£«6O2===6CO2£«6H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½«0.2molMn02ºÍ50mL12mol/LÑÎËá»ìºÏºó¼ÓÈÈ£¬·´Ó¦ÍêÈ«ºóÏòÁôϵÄÈÜÒºÖмÓÈë×ãÁ¿AgNO3ÈÜÒº£¬Éú³ÉAgCl³ÁµíÎïÖʵÄÁ¿Îª£¨²»¿¼ÂÇÑÎËáµÄ»Ó·¢£©£¨ £©

A. µÈÓÚ0.3mol B. СÓÚ0.3mol C. ´óÓÚ 0.3mol£¬Ð¡ÓÚ 0.6mol D. µÈÓÚ 0.6mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÎÒ¹ú¹Å´úµÄ¼¼ÊõÓ¦ÓÃÖУ¬Æ乤×÷Ô­Àí²»Éæ¼°»¯Ñ§·´Ó¦µÄÊÇ

A.ɳÀïÌÔ½ðB.Á¸Ê³Äð¾ÆC.»ðҩʹÓÃD.ÌúµÄÒ±Á¶

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸